To determine where the function \(f\) is increasing, we need to find where \(f'(x) > 0\).

Given \(f'(x) = 3x^2 + 2x - 5\), we set up the inequality:

\(3x^2 + 2x - 5 > 0\).

Factor the quadratic expression:

\((3x + 5)(x - 1) > 0\).

The critical points are found by setting each factor to zero:

\(3x + 5 = 0 \Rightarrow x = -\frac{5}{3}\)

\(x - 1 = 0 \Rightarrow x = 1\)

These critical points divide the number line into intervals. Test each interval to determine where the inequality holds:

• For \(x < -\frac{5}{3}\), choose \(x = -2\): \((3(-2) + 5)((-2) - 1) = (-1)(-3) = 3 > 0\).
• For \(-\frac{5}{3} < x < 1\), choose \(x = 0\): \((3(0) + 5)(0 - 1) = (5)(-1) = -5 < 0\).
• For \(x > 1\), choose \(x = 2\): \((3(2) + 5)(2 - 1) = (11)(1) = 11 > 0\).

Thus, \(f(x)\) is increasing for \(x \leq -\frac{5}{3}\) and \(x > 1\).